Tag: odd and even numbers

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Questions Related to odd and even numbers

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

Let S be a set of all even integers. If the operations:
1. addition 2. subtraction 3. multiplication 4. division
are applied to any pair of numbers from S, then for which operations is the resulting number is S?

  1. $1, 2, 3$ and $4$
  2. $1, 2$ and $3$ only
  3. $1$ and $3$ only
  4. $2$ and $4$ only
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Addition of two even numbers; subtraction of two even numbers and product of two even numbers is an even number.

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

Multiplication of one odd and one even integer is always :

  1. Even

  2. Odd

  3. Can't be determined

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Even integer is an integer having unit digit as a multiple of $2$

So on multiplication with odd integer, unit digit will still remain a multiple of $2$, hence, multiplication of odd and even integer gives even integer.

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

$a, b, c$ are even numbers and $x, y, z$ are odd numbers. Which of the following relationships can't be justified at any cost?
(a) $\dfrac{a \times b}{c} = x \times y$  (b)  $\dfrac{a \times b}{x} = yz$  (c) $\dfrac{xy}{z} = ab$

  1. Only a

  2. Only c

  3. All the three

  4. Only b and c
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Use the rules that 
(a) product of two odd or even numbers are odd and even respectively.
(b) the quotient of the odd number $ \div $ even number or vice-versa may or may not be strictly odd or even.

a. $ ab $ must be a even number and $ \dfrac{ab}{c} $ may be even number or a fraction or odd number and $ xy $ must be a odd number
therefore it can be justified in some case.

b. $ \dfrac { ab }{ x } $ is always a fraction or even number and $ yz $ is always a odd number
therefore it cannot be justified in any case

c.$ \dfrac { xy }{ z } $ maybe odd number or fraction but $ ab $ will always be even.
therefore it cannot be justified.

Relationships given in b and c can't be justified at any cost.
Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

If $f(x)=x^{2}+6x+c$, where $'c'$ is an integer, then $f(0)+f(-1)$ is

  1. an even integer
  2. an odd integer always disable by $3$
  3. an odd integer not divisible by $3$
  4. an odd integer may or not be divisible by $3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$f(0) = c$

$f(-1) = c-5$

$f(0)+f(-1) = 2c-5 = 2(c-3) + 1$

As the above is of the form $2k+1$, it is always odd.

For $c=3$, the above is not divisible by 3 but for $c=4$, it is. Therefore, it may or may not be divisible by 3.

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

If $P$ is an integer between $0$ and $9,R-P=16229$ and $R$ divisible by $11$, then find the value of $\dfrac {P+R-1}{3}$

  1. $5014$
  2. $4514$
  3. $5414$
  4. $5114$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$ R-P = 16229 $
P be b/w $0\& 9 $
and R is divisible by 11
So, $ R = \dfrac{16229+P}{11} $
(and Reminder = 0)
So, $ \Rightarrow (\dfrac{11+P}{11}) $ so $ P = 7 $
and $ R = 16229+7 $
$ = 16236 $
So $ \dfrac{P+R+1}{3} $
$ \Rightarrow \dfrac{16236+7-1}{3} $
$ \Rightarrow \dfrac{16242}{3} = 5414 $ 
Option C is correct 
Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

Consider $n={21}^{52}$, then

  1. number of even divisors of $n$ is $704$
  2. number of odd divisors of $n$ is $2809$
  3. last two digits of $n$ is $41$
  4. number of even divisors of $n$ which are multiple of $9$ is $2705$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

Let,we have


$n = {21^{52}}$

can be written as $n = {\left( {7 \times 3} \right)^{52}}$

$n = {7^{52}}{.3^{52}}$

We know, no. of total divisors of any number

$k = {p^m}.{q^n}$

Total divisors$=(m+1)\,(n+1)$

so, similary here

odd divisors$=(52+1)(52+1)=2809$

Hence the option $(B)$ is correct

But again 

For last two digit

$n = {21^{52}} = {\left( {20 + 1} \right)^{52}}$

${\left( {20 + 1} \right)^{52}}{ = ^{52}}{C _1}{\left( {20} \right)^{52}} + .....{ + ^{52}}{C _{51}}{\left( {20} \right)^1}{ + ^{52}}{C _{52}}{\left( {20} \right)^0}$

For last two digit we notice last two terms 

$=^{52}{C _{51}}\left( {20} \right) + 1$

$ = 52 \times 20 \times 1$

$=1041$

$1041$ has last two digit is $41$

so, option $(C)$ is also correct

Hence both the option $(B)$ and $(C)$ are correct.

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

The smallest odd number formed by using the digits $1,0,3,4$ and $5$ is

  1. $10345$
  2. $10453$
  3. $10543$
  4. $10534$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The smallest odd number using digits $1,0,3,4,5$


$\rightarrow $ We have five digits and we have to make smallest five digit odd numbers.


$\rightarrow$ So, the number cannot start with $0$

$\rightarrow$ For the smallest it should be start with $1$

$\rightarrow$ and second space should be $0$

    $1\\ \overline { 1st } $  $0\\ \overline { 2nd }$  $\;\\ \overline { 3rd } $  $\;\\ \overline { 4rt } $  $\;\\ \overline { 5th } $

$\rightarrow$ Now two space are filled and $3$ are left.

$\rightarrow$ For smallest third place for should be $3$ 

          $\underline { 1 } \underline { 0 } \underline { 3 } \underline {  } \underline {  } $

$\rightarrow $ Now two places are left for and no. should be odd so, last digit should be $5$

So, the number $=10345.$

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

The integer just below $(\sqrt{53}+7)^{11}-2\times 7^{11}$ is 

  1. Divisible by exactly $4$ primes factors
  2. Divisible by exactly $3$ primes factors
  3. is divisible by $7$
  4. has $53$ as its only two digit prime factor
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression (sqrt(53)+7)^11 - 2*7^11 involves binomial expansion. The term (sqrt(53)+7)^11 can be written as (7+sqrt(53))^11 + (7-sqrt(53))^11, which is an integer. The value is approximately 2*7^11, and the integer just below it relates to the properties of these powers.

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

Total number of four digit odd numbers that can be formed using $0,1,2,3,5,7$ are

  1. $192$
  2. $375$
  3. $400$
  4. $720$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

we have the number $0,1,2,3,5,7$

Now the digit should be odd and hence last digit should be 
filled with an odd number 
$ \Rightarrow $ Number of way to filled last number $= 4$ $({\text{i}}{\text{.e}}{\text{. }}1,3,5,7)$
$ \Rightarrow $ Number of way to filled third digit  $= 6$ $({\text{i}}{\text{.e}}{\text{. 0,}}1,2,3,5,7)$
$ \Rightarrow $ Number of way to filled second digit $= 6$ $({\text{i}}{\text{.e}}{\text{. 0,}}1,2,3,5,7)$
$ \Rightarrow $ Number of way to filled first digit $= 5$ $({\text{i}}{\text{.e}.}1,2,3,5,7)$
$ \Rightarrow $ Total 4digit number $=4\times6\times6\times5$
$= 720$
hence,
Opton $D$ is correct answer.